Expected Distance

We can parameterize the circle by $(1 + \cos \theta, 1 + \sin \theta)$ . Hence, the expected value of $D^2$ is $\begin{aligned} \frac{1}{2 \pi} \int_0^{2 \pi} [(1 + \cos \theta)^2 + (1 + \sin \theta)^2] \ d \theta &= \frac{1}{2 \pi} \int_0^{2 \pi} (1 + 2 \cos \theta + \cos^2 \theta + 1 + 2 \sin \theta + \sin^2 \theta) \ d \theta \\ &= \frac{1}{2 \pi} \int_0^{2 \pi} (3 + 2 \cos \theta + 2 \sin \theta) \ d \theta \\ &= 3. \end{aligned}$

More generally, consider a circle centered at $O$ with radius $r$ . Let $P$ be a point, and let $D$ be the distance from $P$ to a random point on the circle. Then the expected value of $D^2$ is $r^2 + OP^2$ .

If $0\leq y\leq 1$ , then for some value of $x$ , the square of the distance can be given by $D_{0\leq y\leq 1}^2=x^2+\left(1-\sqrt{1-(x-1)^2}\right)^2$ .

Similarly, if $1\leq y\leq 2$ , then for some value of $x$ , the square of the distance can be given by $D_{1\leq y\leq 2}^2=x^2+\left(1+\sqrt{1-(x-1)^2}\right)^2$ . Since $y$ has an equal chance of being greater than or less than $1$ for some value of $x$ , we have

$\begin{aligned} \mathbb{E}[D_{0\leq y\leq 2}^2]&=\dfrac{\mathbb{E}[D_{0\leq y\leq 1}^2]+\mathbb{E}[D_{1\leq y\leq 2}^2]}{2}\\ &=\dfrac{1}{2}\left(\dfrac{1}{2}\int_0^2x^2+\left(1-\sqrt{1-(x-1)^2}\right)^2\; dx + \dfrac{1}{2}\int_0^2x^2+\left(1+\sqrt{1-(x-1)^2}\right)^2\; dx\right)\\ &=\dfrac{1}{2}\left(\dfrac{-\pi +6}{2}+\dfrac{\pi +6}{2}\right)\\ &=\boxed{3}. \end{aligned}$

Alternatively, consider that for some point $(x,y)$ on the circle's circumference, we have

$\mathbb{E}[x]=\mathbb{E}[y]=1$

Now the circle has equation $(y-1)^2+(x-1)^2=1$ so that

$D^2=x^2+y^2=2x+2y-1$

and thus

$\mathbb{E}[D^2]=2\mathbb{E}[x+y]-1=2\times 2-1=\boxed{3}.$

Draw a line passing through the origin $O$ and centre $C$ of the circle, (touching the circle at points $A$ and $B$ ), and let $P$ be any point on the circle below this line. Now consider the triangle $\Delta OCP$ with $\angle OCP = \theta$ . Then by the cosine rule

$|OP|^{2} = |OC|^{2} + |CP|^{2} - 2|OC||CP|\cos(\theta) = (\sqrt{2})^{2} + 1^{2} - 2\times \sqrt{2} \times 1 \times \cos(\theta) = 3 - 2\sqrt{2}\cos(\theta)$ .

Now $D = |OP|$ . By symmetry, if we calculate the average of $|OP|^{2}$ over the arc $AB$ , (which has length $\pi$ ), then we will have found the expected value of $D^{2}$ . To do this we need to take the integral of $|OP|^{2}$ with respect to $\theta$ as $\theta$ goes from $0$ to $\pi$ and then divide by the length of arc $AB$ :

$D^{2} = \dfrac{1}{\pi} \displaystyle \int_{0}^{\pi} (3 - 2\sqrt{2}\cos(\theta)) d\theta = \dfrac{1}{\pi} \left(3\theta - 2\sqrt{2}\sin(\theta)\right)_{0}^{\pi} = \dfrac{1}{\pi} \times 3\pi = \boxed{3}$ .

The answer is immediate from a physics point of view, since we are looking for the radius of gyration of a circumference taken from origin in our case. The Steiner formula gives us that value of Moment of Inertia of a plane figure from an arbitrary point will be Moment of inertia from center of gravity of the figure plus the mass of figure times the square of distance between the given point and center of gravity. Dividing the resulting Moment of Inertia by the mass 2piR we get the radius of gyration squared must be R + 2R. Being R= 1 so D^2=3. We can also tackle the problem from a more mathematical path using polarity.

In the polar coordinate, the coordinates of any point on the circle is of the form $\big(1+\cos(\theta), 1+\sin(\theta)\big),$ where $\theta$ is a uniformly distributed random variable in the range $[0, 2\pi]$ . Thus, $D^2= (1+\cos(\theta))^2+ (1+\sin(\theta))^2=3+2(\cos(\theta) + \sin (\theta)).$ Taking expectations of both sides and realizing that, under the given uniform distribution $\mathbb{E}(\cos(\theta))=\mathbb{E}(\sin(\theta))=0$ , we conclude that $\mathbb{E}(D^2)=3.$